Left Termination of the query pattern tc(b,f) w.r.t. the given Prolog program could successfully be proven:



↳ PROLOG
  ↳ PrologToPiTRSProof

p2(a0, b0).
p2(b0, c0).
tc2(X, X).
tc2(X, Y) :- p2(X, Z), tc2(Z, Y).


With regard to the inferred argument filtering the predicates were used in the following modes:
tc2: (b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


tc_2_in_ga2(X, X) -> tc_2_out_ga2(X, X)
tc_2_in_ga2(X, Y) -> if_tc_2_in_1_ga3(X, Y, p_2_in_ga2(X, Z))
p_2_in_ga2(a_0, b_0) -> p_2_out_ga2(a_0, b_0)
p_2_in_ga2(b_0, c_0) -> p_2_out_ga2(b_0, c_0)
if_tc_2_in_1_ga3(X, Y, p_2_out_ga2(X, Z)) -> if_tc_2_in_2_ga4(X, Y, Z, tc_2_in_ga2(Z, Y))
if_tc_2_in_2_ga4(X, Y, Z, tc_2_out_ga2(Z, Y)) -> tc_2_out_ga2(X, Y)

The argument filtering Pi contains the following mapping:
tc_2_in_ga2(x1, x2)  =  tc_2_in_ga1(x1)
a_0  =  a_0
b_0  =  b_0
c_0  =  c_0
tc_2_out_ga2(x1, x2)  =  tc_2_out_ga1(x2)
if_tc_2_in_1_ga3(x1, x2, x3)  =  if_tc_2_in_1_ga1(x3)
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_tc_2_in_2_ga4(x1, x2, x3, x4)  =  if_tc_2_in_2_ga1(x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

tc_2_in_ga2(X, X) -> tc_2_out_ga2(X, X)
tc_2_in_ga2(X, Y) -> if_tc_2_in_1_ga3(X, Y, p_2_in_ga2(X, Z))
p_2_in_ga2(a_0, b_0) -> p_2_out_ga2(a_0, b_0)
p_2_in_ga2(b_0, c_0) -> p_2_out_ga2(b_0, c_0)
if_tc_2_in_1_ga3(X, Y, p_2_out_ga2(X, Z)) -> if_tc_2_in_2_ga4(X, Y, Z, tc_2_in_ga2(Z, Y))
if_tc_2_in_2_ga4(X, Y, Z, tc_2_out_ga2(Z, Y)) -> tc_2_out_ga2(X, Y)

The argument filtering Pi contains the following mapping:
tc_2_in_ga2(x1, x2)  =  tc_2_in_ga1(x1)
a_0  =  a_0
b_0  =  b_0
c_0  =  c_0
tc_2_out_ga2(x1, x2)  =  tc_2_out_ga1(x2)
if_tc_2_in_1_ga3(x1, x2, x3)  =  if_tc_2_in_1_ga1(x3)
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_tc_2_in_2_ga4(x1, x2, x3, x4)  =  if_tc_2_in_2_ga1(x4)


Pi DP problem:
The TRS P consists of the following rules:

TC_2_IN_GA2(X, Y) -> IF_TC_2_IN_1_GA3(X, Y, p_2_in_ga2(X, Z))
TC_2_IN_GA2(X, Y) -> P_2_IN_GA2(X, Z)
IF_TC_2_IN_1_GA3(X, Y, p_2_out_ga2(X, Z)) -> IF_TC_2_IN_2_GA4(X, Y, Z, tc_2_in_ga2(Z, Y))
IF_TC_2_IN_1_GA3(X, Y, p_2_out_ga2(X, Z)) -> TC_2_IN_GA2(Z, Y)

The TRS R consists of the following rules:

tc_2_in_ga2(X, X) -> tc_2_out_ga2(X, X)
tc_2_in_ga2(X, Y) -> if_tc_2_in_1_ga3(X, Y, p_2_in_ga2(X, Z))
p_2_in_ga2(a_0, b_0) -> p_2_out_ga2(a_0, b_0)
p_2_in_ga2(b_0, c_0) -> p_2_out_ga2(b_0, c_0)
if_tc_2_in_1_ga3(X, Y, p_2_out_ga2(X, Z)) -> if_tc_2_in_2_ga4(X, Y, Z, tc_2_in_ga2(Z, Y))
if_tc_2_in_2_ga4(X, Y, Z, tc_2_out_ga2(Z, Y)) -> tc_2_out_ga2(X, Y)

The argument filtering Pi contains the following mapping:
tc_2_in_ga2(x1, x2)  =  tc_2_in_ga1(x1)
a_0  =  a_0
b_0  =  b_0
c_0  =  c_0
tc_2_out_ga2(x1, x2)  =  tc_2_out_ga1(x2)
if_tc_2_in_1_ga3(x1, x2, x3)  =  if_tc_2_in_1_ga1(x3)
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_tc_2_in_2_ga4(x1, x2, x3, x4)  =  if_tc_2_in_2_ga1(x4)
TC_2_IN_GA2(x1, x2)  =  TC_2_IN_GA1(x1)
IF_TC_2_IN_2_GA4(x1, x2, x3, x4)  =  IF_TC_2_IN_2_GA1(x4)
P_2_IN_GA2(x1, x2)  =  P_2_IN_GA1(x1)
IF_TC_2_IN_1_GA3(x1, x2, x3)  =  IF_TC_2_IN_1_GA1(x3)

We have to consider all (P,R,Pi)-chains

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

TC_2_IN_GA2(X, Y) -> IF_TC_2_IN_1_GA3(X, Y, p_2_in_ga2(X, Z))
TC_2_IN_GA2(X, Y) -> P_2_IN_GA2(X, Z)
IF_TC_2_IN_1_GA3(X, Y, p_2_out_ga2(X, Z)) -> IF_TC_2_IN_2_GA4(X, Y, Z, tc_2_in_ga2(Z, Y))
IF_TC_2_IN_1_GA3(X, Y, p_2_out_ga2(X, Z)) -> TC_2_IN_GA2(Z, Y)

The TRS R consists of the following rules:

tc_2_in_ga2(X, X) -> tc_2_out_ga2(X, X)
tc_2_in_ga2(X, Y) -> if_tc_2_in_1_ga3(X, Y, p_2_in_ga2(X, Z))
p_2_in_ga2(a_0, b_0) -> p_2_out_ga2(a_0, b_0)
p_2_in_ga2(b_0, c_0) -> p_2_out_ga2(b_0, c_0)
if_tc_2_in_1_ga3(X, Y, p_2_out_ga2(X, Z)) -> if_tc_2_in_2_ga4(X, Y, Z, tc_2_in_ga2(Z, Y))
if_tc_2_in_2_ga4(X, Y, Z, tc_2_out_ga2(Z, Y)) -> tc_2_out_ga2(X, Y)

The argument filtering Pi contains the following mapping:
tc_2_in_ga2(x1, x2)  =  tc_2_in_ga1(x1)
a_0  =  a_0
b_0  =  b_0
c_0  =  c_0
tc_2_out_ga2(x1, x2)  =  tc_2_out_ga1(x2)
if_tc_2_in_1_ga3(x1, x2, x3)  =  if_tc_2_in_1_ga1(x3)
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_tc_2_in_2_ga4(x1, x2, x3, x4)  =  if_tc_2_in_2_ga1(x4)
TC_2_IN_GA2(x1, x2)  =  TC_2_IN_GA1(x1)
IF_TC_2_IN_2_GA4(x1, x2, x3, x4)  =  IF_TC_2_IN_2_GA1(x4)
P_2_IN_GA2(x1, x2)  =  P_2_IN_GA1(x1)
IF_TC_2_IN_1_GA3(x1, x2, x3)  =  IF_TC_2_IN_1_GA1(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

Pi DP problem:
The TRS P consists of the following rules:

TC_2_IN_GA2(X, Y) -> IF_TC_2_IN_1_GA3(X, Y, p_2_in_ga2(X, Z))
IF_TC_2_IN_1_GA3(X, Y, p_2_out_ga2(X, Z)) -> TC_2_IN_GA2(Z, Y)

The TRS R consists of the following rules:

tc_2_in_ga2(X, X) -> tc_2_out_ga2(X, X)
tc_2_in_ga2(X, Y) -> if_tc_2_in_1_ga3(X, Y, p_2_in_ga2(X, Z))
p_2_in_ga2(a_0, b_0) -> p_2_out_ga2(a_0, b_0)
p_2_in_ga2(b_0, c_0) -> p_2_out_ga2(b_0, c_0)
if_tc_2_in_1_ga3(X, Y, p_2_out_ga2(X, Z)) -> if_tc_2_in_2_ga4(X, Y, Z, tc_2_in_ga2(Z, Y))
if_tc_2_in_2_ga4(X, Y, Z, tc_2_out_ga2(Z, Y)) -> tc_2_out_ga2(X, Y)

The argument filtering Pi contains the following mapping:
tc_2_in_ga2(x1, x2)  =  tc_2_in_ga1(x1)
a_0  =  a_0
b_0  =  b_0
c_0  =  c_0
tc_2_out_ga2(x1, x2)  =  tc_2_out_ga1(x2)
if_tc_2_in_1_ga3(x1, x2, x3)  =  if_tc_2_in_1_ga1(x3)
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_tc_2_in_2_ga4(x1, x2, x3, x4)  =  if_tc_2_in_2_ga1(x4)
TC_2_IN_GA2(x1, x2)  =  TC_2_IN_GA1(x1)
IF_TC_2_IN_1_GA3(x1, x2, x3)  =  IF_TC_2_IN_1_GA1(x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

TC_2_IN_GA2(X, Y) -> IF_TC_2_IN_1_GA3(X, Y, p_2_in_ga2(X, Z))
IF_TC_2_IN_1_GA3(X, Y, p_2_out_ga2(X, Z)) -> TC_2_IN_GA2(Z, Y)

The TRS R consists of the following rules:

p_2_in_ga2(a_0, b_0) -> p_2_out_ga2(a_0, b_0)
p_2_in_ga2(b_0, c_0) -> p_2_out_ga2(b_0, c_0)

The argument filtering Pi contains the following mapping:
a_0  =  a_0
b_0  =  b_0
c_0  =  c_0
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
TC_2_IN_GA2(x1, x2)  =  TC_2_IN_GA1(x1)
IF_TC_2_IN_1_GA3(x1, x2, x3)  =  IF_TC_2_IN_1_GA1(x3)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ RuleRemovalProof

Q DP problem:
The TRS P consists of the following rules:

TC_2_IN_GA1(X) -> IF_TC_2_IN_1_GA1(p_2_in_ga1(X))
IF_TC_2_IN_1_GA1(p_2_out_ga1(Z)) -> TC_2_IN_GA1(Z)

The TRS R consists of the following rules:

p_2_in_ga1(a_0) -> p_2_out_ga1(b_0)
p_2_in_ga1(b_0) -> p_2_out_ga1(c_0)

The set Q consists of the following terms:

p_2_in_ga1(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_TC_2_IN_1_GA1, TC_2_IN_GA1}.
By using a polynomial ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

IF_TC_2_IN_1_GA1(p_2_out_ga1(Z)) -> TC_2_IN_GA1(Z)

Strictly oriented rules of the TRS R:

p_2_in_ga1(a_0) -> p_2_out_ga1(b_0)
p_2_in_ga1(b_0) -> p_2_out_ga1(c_0)

Used ordering: POLO with Polynomial interpretation:

POL(p_2_out_ga1(x1)) = 1 + 2·x1   
POL(a_0) = 2   
POL(IF_TC_2_IN_1_GA1(x1)) = x1   
POL(c_0) = 0   
POL(TC_2_IN_GA1(x1)) = 2·x1   
POL(p_2_in_ga1(x1)) = 2·x1   
POL(b_0) = 1   



↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

TC_2_IN_GA1(X) -> IF_TC_2_IN_1_GA1(p_2_in_ga1(X))

R is empty.
The set Q consists of the following terms:

p_2_in_ga1(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_TC_2_IN_1_GA1, TC_2_IN_GA1}.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.